If $G$ is a group with subgroup $H$, then we have the restriction functor $\operatorname{Res}$ from $G-\operatorname{mod}$ to $H-\operatorname{mod}$. We also have this idea of induction, a functor $\operatorname{Ind}^G_H$ from $H-\operatorname{mod}$ to $G-\operatorname{mod}$. These are adjoints, which means (I think) that $\operatorname{Hom}_G(\operatorname{Ind}^G_H(V), U) \cong \operatorname{Hom}_H(V,\operatorname{Res}(U))$ naturally, for $G$-modules $U$ and $H$-modules $V$.

For locally compact groups, there is a theory worked out by MacKey and others. Actually, I have only read Rieffel's work on the subject (as I come from a functional Analysis background). For a locally compact $G$ and closed subgroup $H$, there is a very satisfactory notion of the functor $\operatorname{Ind}^G_H$ (where we consider "Hermitian modules", i.e. unitary representations on Hilbert spaces). What I don't see is how (or even if) this relates to the restriction functor?

In the topological setting, are $\operatorname{Ind}^G_H$ and $\operatorname{Res}$ in any sense adjoints?

A slightly vague rider-- if (as I suspect) the answer is "no", can we be more precise about why the answer is no?

I think induction in this setting is actually conduction with topological overtones, so it maybe right adjoint to restriction but I am no expert.
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Benjamin SteinbergDec 13 '11 at 17:26

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@Benjamin: If I understand your comment, then I should have stressed right adjoint-- I think it's too much to hope that $\operatorname{Hom}_G(U,\operatorname{Ind}^G_H(V)) \cong \operatorname{Hom}_H(\operatorname{Res}(U),V)$ I think. So, yes, I agree!
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Matthew DawsDec 13 '11 at 20:28

I'm a bit confused, Matt: the isomorphism written in your question says that induction is left adjoint to restriction; the induced representation is the one "freely generated" by a given representation of $H$. The isomorphism in your comment would be saying that induction is right adjoint to restriction.
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Yemon ChoiDec 13 '11 at 23:52

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@Yemon: Yes, Induction is a left adjoint for Restriction; so Restriction is a right adjoint for Induction (I think I didn't commit myself as to which was which, as it were!) As for which modules, I already say "'hermitian module', i.e. unitary representations on Hilbert spaces"...??
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Matthew DawsDec 14 '11 at 8:24

2 Answers
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If $G$ is compact this is the Frobenius Reciprocity Theorem, see e.g., Section 6.2 in Folland's A Course in Abstract Harmonic Analysis for a proof. When $G$ is not compact then this fails already for $H$ the trivial group and $U$, and $V$ trivial representations. Indeed, in this case Ind$_H^G(1_H) = L^2G$ the left regular representation, and since constant functions are not square integrable we have Hom$_G( L^2G, 1_G ) = \{ 0 \}$, while Hom$_H(1_H, 1_H) \not= \{ 0 \}$.

Ah, so the answer is "no" for a pretty simple reason! I was holding out for some suggestions about when we get a positive result; but none was forthcoming. So I guess MO etiquette says I accept this (as it nicely answers my main question) and, ahem, do some thinking of my own...
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Matthew DawsDec 14 '11 at 20:14

In fact the right relation is
$$ Hom_G ( - , Ind_H^G - ) = Hom_H( Res - , -).$$
For compact group, it does not really matter by the Peter Weyl theorem, but it is essential as soon as you omit compactness.

I want to add that Frobenius reciprocity usually boils down to tensor-hom adjointness, so essentailly the nuclearity of the module category is necessary. For a positive answer, you have to be more restrictive.

For Mackey's theory of induced representation and its variants of Frobenius reciprocity, have a look at Baruk&Raczka "Theory of group representations and applications" pg. 549.

For the locally profinite groups, there is a really nice treatment in Bushnell-Henniart "Local Langlands on GL(2)" on pg. 18ff.

To add upon Jesse Peterson's example: If $G$ is locally compact, then
$$ Hom_G( L^1 G , 1_G) \cong \mathbb{C} \cong End_H(1_H).$$
(Hint: The Haar integraal is the only element on the left hand side.)
So the category of Hilbert spaces is not for all purposes the prefered one, you want to work in.